SOLUTION: Rohan mixed a 20% acid solution with an 80% acids solution to obtain 120 liters of a solution that is 50% acid. How much of each solution must be used to achieve the desired mixtur

Question 1076131: Rohan mixed a 20% acid solution with an 80% acids solution to obtain 120 liters of a solution that is 50% acid. How much of each solution must be used to achieve the desired mixture?
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52863)   (Show Source): You can put this solution on YOUR website!
.

The equation is 0.2*x + 0.8*(120-x) = 0.5*120, where "x" is the volume of the 20% acid solution. Simplify and solve for x: 0.2x + 96 -0.8x = 60, -0.6x = 60 - 96, -0.6x = - 36, x = = 60. Answer. 60 liters of the 20% solution and 120-60 = 60 liters of the 80% solution.

Solved.

There is entire bunch of lessons covering various types of mixture problems
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
in this site.

Read them and become an expert in solution the mixture word problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
50% is the average of 20% and 80%.
--> equal amounts.
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